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Dahasolnce [82]
3 years ago
7

3x=12-2y write in standard form

Mathematics
1 answer:
klio [65]3 years ago
6 0
Standard form is y = mx + b.

So,

3x = 12 - 2y
+ 2y     + 2y

3x + 2y = 12
-3x         - 3x

2y = -3x + 12
---     ----   ----
 2       2      2

y = (-3/2)x + 6
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avanturin [10]

Answer:

<h2>Pythagorean Theorem: a^2 + b^2 = c^2</h2>

But what is the actual question?

4 0
3 years ago
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Solve the following 19+n=30​
Sliva [168]

Answer:

n=11

Step-by-step explanation:

19+n=30

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2.The number of grams A of a certain radioactive substance present at time, in yearsfrom the present, t is given by the formulaA
professor190 [17]

Answer:

Given that,

The number of grams A of a certain radioactive substance present at time, in years

from the present, t is given by the formula

A=45e^{-0.0045(t)}

a) To find the initial amount of this substance

At t=0, we get

A=45e^{-0.0045(0)}A=45e^0

We know that e^0=1 ( anything to the power zero is 1)

we get,

A=45

The initial amount of the substance is 45 grams

b)To find thehalf-life of this substance

To find t when the substance becames half the amount.

A=45/2

Substitute this we get,

\frac{45}{2}=45e^{-0.0045(t)}

\frac{1}{2}=e^{-0.0045(t)}

Taking natural logarithm on both sides we get,

\ln (\frac{1}{2})=-0.0045(t)^{}(-1)\ln (\frac{1}{2})=0.0045(t)\ln (\frac{1}{2})^{-1}=0.0045(t)\ln (2)=0.0045(t)0.6931=0.0045(t)t=\frac{0.6931}{0.0045}t=154.02

Half-life of this substance is 154.02

c) To find the amount of substance will be present around in 2500 years

Put t=2500

we get,

A=45e^{-0.0045(2500)}A=45e^{-11.25}A=45\times0.000013=0.000585A=0.000585

The amount of substance will be present around in 2500 years is 0.000585 grams

4 0
11 months ago
(-5)+(+7)=( ____) ???
Anni [7]

Answer: 2

Step-by-step explanation:

7 0
3 years ago
How can probabilities be used to make fair decisions?
GarryVolchara [31]

he magician starts with the birthday boy and moves clockwise, passing out 100100100100 pieces of paper numbered 1111 through 100100100100. He cycles around the circle until all the pieces are distributed. He then uses a random number generator to pick an integer 1111 through 100100100100, and chooses the volunteer with that number.

Method2: The magician starts with the birthday boy and moves counter-clockwise, passing out 75757575 pieces of paper numbered 1111 through 75757575. He cycles around the circle until all the pieces are distributed. He then uses a random number generator to pick an integer 1111 through 75757575, and chooses the volunteer with that number.

Method 3\: The magician starts with the birthday boy and moves clockwise, passing out 30303030 pieces of paper numbered 1111 through 30303030. He cycles around the circle until all the pieces are distributed. He gives #1111 to the birthday boy, #2222 to the next kid, and so on. He then counts the number of windows in the room and chooses the volunteer with that number.

yes probabilites can be used to make fair ones

thanx

heya

6 0
3 years ago
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